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I actually put quite a bit of effort into this but couldn't get a rigorous proof of it, which honestly pissed me off. Here's an example that I used to put me on the right track:
sin(pi/2)=1
Check a=b=pi/4 (just kind of a "duh" example, but one that should definitely work if the condition is true.)
Then (sqrt(2)/2)^2+(sqrt(2)/2)^2=1/2+1/2=1, so that example works.
Here's some identities you might be able to use.
sin(a)=cos(pi/2-a)
cos(a)=sin(pi/2-a)
cos(a)=cos(-a) (and so cos(pi/2-a)=cos(a-pi/2)=sin(a))
sin(a)=-sin(-a)
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
sin^2(a)=(1-cos(2a))/2
cos^2(a)=(1+cos(2a))/2)
sin^2(a)+cos^2(a)=1

I rather hope this isn't due tomorrow, because I'd like to be able to come back to it once I have some time, if only for the challenge.

I actually put quite a bit of effort into this but couldn't get a rigorous proof of it, which honestly pissed me off. Here's an example that I used to put me on the right track:
sin(pi/2)=1
Check a=b=pi/4
Then (sqrt(2)/2)^2+(sqrt(2)/2)^2=1/2+1/2=1, so that example works.
Here's some identities you might be able to use.
sin(a)=cos(pi/2-a)
cos(a)=sin(pi/2-a)
cos(a)=cos(-a) (and so cos(pi/2-a)=cos(a-pi/2)=sin(a))
sin(a)=-sin(-a)
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
sin^2(a)=(1-cos(2a))/2
cos^2(a)=(1+cos(2a))/2)
sin^2(a)+cos^2(a)=1

I rather hope this isn't due tomorrow, because I'd like to be able to come back to it once I have some time, if only for the challenge.

Let him do his own work.

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At the moment, that's what I'm doing...that's just a bunch of identities that I considered as I tried to approach the problem.

Wow this is actually way simpler than I thought, here you go:
sin^2(a)+sin^2(b) = sin(a)cos(b)+cos(a)sin(b) (identity listed above)
Hence:
sin(a)=cos(b)
cos(a)=sin(b)
(Look at the equation for a second it should be pretty clear).
Because sin(a)=cos(pi/2-a) (again identity listed above), b=pi/2-a. Add a to both sides and you show that a+b=pi/2. Alternatively, cos(a)=sin(b), and cos(a)=sin(pi/2-a), so then a=pi/2-b, and by the same reasoning a+b=pi/2. Q.E.D.

Trig really comes down to knowing the identities and not making the problem too complicated. If you know them, you can usually make it through any problem unscathed.

I actually put quite a bit of effort into this but couldn't get a rigorous proof of it, which honestly pissed me off. Here's an example that I used to put me on the right track:
sin(pi/2)=1
Check a=b=pi/4 (just kind of a "duh" example, but one that should definitely work if the condition is true.)
Then (sqrt(2)/2)^2+(sqrt(2)/2)^2=1/2+1/2=1, so that example works.
Here's some identities you might be able to use.
sin(a)=cos(pi/2-a)
cos(a)=sin(pi/2-a)
cos(a)=cos(-a) (and so cos(pi/2-a)=cos(a-pi/2)=sin(a))
sin(a)=-sin(-a)
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
sin^2(a)=(1-cos(2a))/2
cos^2(a)=(1+cos(2a))/2)
sin^2(a)+cos^2(a)=1

I rather hope this isn't due tomorrow, because I'd like to be able to come back to it once I have some time, if only for the challenge.

Originally Posted by alfaroverall

At the moment, that's what I'm doing...that's just a bunch of identities that I considered as I tried to approach the problem.

Wow this is actually way simpler than I thought, here you go:
sin^2(a)+sin^2(b) = sin(a)cos(b)+cos(a)sin(b) (identity listed above)
Hence:
sin(a)=cos(b)
cos(a)=sin(b)
(Look at the equation for a second it should be pretty clear).
Because sin(a)=cos(pi/2-a) (again identity listed above), b=pi/2-a. Add a to both sides and you show that a+b=pi/2. Alternatively, cos(a)=sin(b), and cos(a)=sin(pi/2-a), so then a=pi/2-b, and by the same reasoning a+b=pi/2. Q.E.D.

Trig really comes down to knowing the identities and not making the problem too complicated. If you know them, you can usually make it through any problem unscathed.

Lol, when I first saw the OP I was thinking "alfaroverall is solving that problem right now"

At the moment, that's what I'm doing...that's just a bunch of identities that I considered as I tried to approach the problem.

Wow this is actually way simpler than I thought, here you go:
sin^2(a)+sin^2(b) = sin(a)cos(b)+cos(a)sin(b) (identity listed above)
Hence:
sin(a)=cos(b)
cos(a)=sin(b)
(Look at the equation for a second it should be pretty clear).
Because sin(a)=cos(pi/2-a) (again identity listed above), b=pi/2-a. Add a to both sides and you show that a+b=pi/2. Alternatively, cos(a)=sin(b), and cos(a)=sin(pi/2-a), so then a=pi/2-b, and by the same reasoning a+b=pi/2. Q.E.D.

Trig really comes down to knowing the identities and not making the problem too complicated. If you know them, you can usually make it through any problem unscathed.

I haven't done any of this for a while. Will you explain how you know sin a = cos b?

I haven't done any of this for a while. Will you explain how you know sin a = cos b?

In the first term on the left and right side, you have a sin(a) factor. In the second term on the left and right hand side, you have a sin(b) factor. But in the first one on the right hand side, the second sin(a) factor is replaced by a cos(b) factor, while in the second one on the right hand site, the second sin(b) factor is replaced by a cos(a) factor. Those second factors have to be equal to the factors they "replaced" in order for the equation to turn out equal.

It's a little easier to understand if you think of it this way. If sin^2(a)=sin(a)cos(b) and simultaneously sin^2(b)=cos(a)sin(b), then the relationship we're looking at is true. (In short, if a=c and b=d, then a+b=c+d.) So if you check each relationship, you can do some routine algebra to find what I showed above:
sin^2(a)=sin(a)cos(b)
(divide both sides by sin(a))
sin(a)=cos(b)
Similarly:
sin^2(b)=cos(a)sin(b)
sin(b)=cos(a)
These are then only BOTH true (with both a and b being acute) if a+b=pi/2, by what I showed above.